[Solved] for(++i;++i;++i) the second argument is < or

Question

In the second argument it is actually ++i!=0, The loop is interpreted as

for(++i;++i!=0;++i)

If you start with a positive i or 0, it will be an infinite loop and will invoke undefined behavior when i reaches INT_MAX.

If i was -Ve initially the loop may stop at a defined run.

EDIT: As you changed your question, Your code will not crash, but you can clearly understand the dry-run by replacing the second ++i with ++i!=0.

So the 1st iteration becomes:

(i=1;2!=0;++i/*this will execute later*/)

2nd iteration becomes:

i=3 //this is from the 1st iteration last part.
(/*initialization is done 1st time only*/;4!=0;++i/*again, this will execute after this iteration*/)

Accepted Answer